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# NCERT Solutions for Class 8 Maths Exercise 8.3

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NCERT solutions for Class 8 Maths Comparing Quantities

## NCERT Solutions for Class 8 Maths Comparing Quantities

###### 1. Calculate the amount and compound interest on:

(a) Rs.10,800 for 3 years at  per annum compounded annually.

(b) Rs.18,000 for  years at 10% per annum compounded annually.

(c) Rs.62,500 for  years at 8% per annum compounded annually.

(d) Rs.8,000 for  years at 9% per annum compounded half yearly. (You could the year by year calculation using S.I. formula to verify).

(e) Rs.10,000 for  years at 8% per annum compounded half yearly.

Ans. (a) Here, Principal (P) = Rs. 10800, Time  = 3 years,

Rate of interest (R) =

Amount (A) =

=  =

=  =

=

= Rs. 15,377.34

Compound Interest (C.I.) = A – P

= Rs. 10800 – Rs. 15377.34 = Rs. 4,577.34

###### (b) Here, Principal (P) = Rs. 18,000, Time  =   years, Rate of interest (R)

= 10% p.a.

Amount (A) =

=  =

=  =

= Rs. 21,780

Interest for  years on Rs. 21,780 at rate of 10% =  = Rs. 1,089

Total amount for  years

= Rs. 21,780 + Rs. 1089 = Rs. 22,869

Compound Interest (C.I.) = A – P

= Rs. 22869 – Rs. 18000 = Rs. 4,869

###### (c) Here, Principal (P) = Rs. 62500, Time  = =  years = 3 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 70,304

Compound Interest (C.I.) = A – P

= Rs. 70304 – Rs. 62500 = Rs. 7,804

###### (d) Here, Principal (P) = Rs. 8000, Time  = 1 years = 2 years(compounded half yearly)

Rate of interest (R) = 9% =  (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 8,736.20

Compound Interest (C.I.) = A – P

= Rs. 8736.20 – Rs. 8000

= Rs. 736.20

###### (e) Here, Principal (P) = Rs. 10,000, Time  = 1 years = 2 years (compounded half yearly)

Rate of interest (R) = 8% = 4% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 10,816

Compound Interest (C.I.) = A – P

= Rs. 10,816 – Rs. 10,000 = Rs. 816

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 2. Kamala borrowed Rs.26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for  years).

Ans. Here, Principal (P) = Rs. 26,400, Time  = 2 years 4 months, Rate of interest (R) = 15% p.a.

Amount for 2 years (A) =

=  =

=  =

= Rs. 34,914

Interest for 4 months =  years at the rate of 15% =

= Rs. 1745.70

Total amount = Rs. 34,914 + Rs. 1,745.70

= Rs. 36,659.70

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 3. Fabina borrows Rs.12,500 per annum for 3 years at simple interest and Radhaborrows the same amount for the same time period at 10% per annnum, compounded annually. Who pays more interest and by how much?

Ans. Here, Principal (P) = Rs.12,500, Time  = 3 years, Rate of interest (R)

= 12% p.a.

Simple Interest for Fabina =

=  = Rs. 4,500

Amount for Radha, P = Rs. 12,500, R = 10% and  = 3 years

Amount (A)  =

= =

= =

= Rs. 16,637.50

C.I. for Radha = A – P

= Rs. 16,637.50 – Rs. 12,500 = Rs. 4,137.50

Here, Fabina pays more interest

= Rs. 4,500 – Rs. 4,137.50 = Rs. 362.50

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 4. I borrowsRs.12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

Ans. Here, Principal (P) = Rs.12,000, Time  = 2 years, Rate of interest (R) = 6% p.a.

Simple Interest =

=  = Rs. 1,440

Had he borrowed this sum at 6% p.a., then

Compound Interest  =

=

=

=

=

= Rs. 13,483.20 – Rs. 12,000

= Rs. 1,483.20

Difference in both interests

= Rs. 1,483.20 – Rs. 1,440.00 = Rs. 43.20

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 5. Vasudevan invested Rs.60,000 at an interest rate of 12% per ann^um compounded half yearly. What amount would he get:

(i) after 6 months?

(ii) after 1 year?

Ans. (i) Here, Principal (P) = Rs. 60,000,

Time  = 6 months = 1 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 63,600

After 6 months Vasudevan would get amount Rs. 63,600.

(ii) Here, Principal (P) = Rs. 60,000,

Time  = 1 year = 2 year(compounded half yearly)

Rate of interest (R) = 12% = 6% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 67,416

After 1 year Vasudevan would get amount Rs. 67,416.

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 6. Arif took a loan of Rs.80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after  years if the interest is:

(i) compounded annually.

(ii) compounded half yearly.

Ans. (i) Here, Principal (P) = Rs. 80,000, Time  =  years, Rate of interest (R) = 10%

Amount for 1 year (A)  =

=

=

=

= Rs. 88,000

Interest for  year =

= Rs. 4,400

Total amount = Rs. 88,000 + Rs. 4,400 = Rs. 92,400

(ii) Here, Principal (P) = Rs.80,000,

Time  =  year = 3year (compounded half yearly)

Rate of interest (R) = 10% = 5% (compounded half yearly)

Amount (A) =

=

=

=

=

= Rs. 92,610

Difference in amounts

= Rs. 92,610 – Rs. 92,400 = Rs. 210

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 7. Maria invested Rs.8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find:

(i) The amount credited against her name at the end of the second year.

(ii) The interest for the third year.

Ans. (i) Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time  = 2 years

Amount (A)  =

=

=

=

=

= Rs. 8,820

###### (ii) Here, Principal (P) = Rs. 8000, Rate of Interest (R) = 5%, Time  = 3 years

Amount (A)  =

=

=

=

=

= Rs. 9,261

Interest for 3rd year = A – P

= Rs. 9,261 – Rs. 8,820 = Rs. 441

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 8. Find the amount and the compound interest on Rs.10,000 for  years at 10% per annum, compounded half yearly.

Would this interest be more than the interest he would get if it was compounded annually?

Ans. Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10% = 5% (compounded half yearly)

Time  =  years = 3 years (compounded half yearly)

Amount (A)  =

=

=

=

=

= Rs. 11,576.25

Compound Interest (C.I.) = A – P

= Rs. 11,576.25 – Rs. 10,000 = Rs. 1,576.25

If it is compounded annually, then

Here, Principal (P) = Rs. 10000, Rate of Interest (R) = 10%, Time  =  years

Amount (A) for 1 year  =

=

=

=

=

= Rs. 11,000

Interest for  year =  = Rs. 550

Total amount = Rs. 11,000 + Rs. 550

= Rs. 11,550

Now, C.I. = A – P = Rs. 11,550 – Rs. 10,000

= Rs. 1,550

Yes, interest Rs. 1,576.25 is more than Rs. 1,550.

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 9. Find the amount which Ram will get on Rs.4,096, if he gave it for 18 months at  per annum, interest being compounded half yearly.

Ans. Here, Principal (P) = Rs. 4096,

Rate of Interest (R) =

=  (compounded half yearly)

Time  = 18 months =  years = 3 years (compounded half yearly)

Amount (A)  =

=

=

=

=

= Rs. 4,913

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.

(i) Find the population in 2001.

(ii) What would be its population in 2005?

Ans. (i) Here,  A2003 = Rs. 54,000, R = 5%,  = 2 years

Population would be less in 2001 than 2003 in two years.

Here population is increasing.

A2003 =

54000 =

54000 =

54000 =

54000 =

48,980 (approx.)

###### (ii) According to question, population is increasing. Therefore population in 2005,

A2005  =

=

=

=

=

= 59,535

Hence population in 2005 would be 59,535.

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 11. In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.

Ans. Here, Principal (P) = 5,06,000, Rate of Interest (R) = 2.5%, Time  = 2 hours

After 2 hours, number of bacteria,

Amount (A)  =

=

=

=

=

=

= 5,31,616.25

Hence, number of bacteria after two hours are 531616 (approx.).

NCERT Solutions for Class 8 Maths Exercise 8.3

###### 12. A scooter was bought at Rs.42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Ans. Here, Principal (P) = Rs. 42,000, Rate of Interest (R) = 8%, Time  = 1 years

Amount (A)  =

=

=

=

=

= Rs. 38,640

Hence, the value of scooter after one year is Rs. 38,640.

## NCERT Solutions for Class 8 Maths Exercise 8.3

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### 34 thoughts on “NCERT Solutions for Class 8 Maths Exercise 8.3”

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